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Drupady [299]
3 years ago
9

Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

ght travels in one year. State the speed in mi/s, using scientific notation. (1 mile = 5280 feet exactly, by definition; i.e., this is not a measurement.) A. 1.8600 x 105 mi/s Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance light travels in one year.
State the speed in mi/s, using scientific notation. (1 mile = 5280 feet exactly, by definition; i.e., this is not a measurement.)

A. 1.8600 x 105 mi/s
B. 5.1854 x 1012 mi/s
Physics
1 answer:
larisa86 [58]3 years ago
6 0

A. 1.8600 x 105 mi/s

Explanation:

The problem gives us the value of the speed of light in ft/s: v=982,080,000 ft/s

We know the conversion factor between feet and miles:

1 mi = 5280 ft

So we can apply this conversion factor to the speed of light:

v=982,080,000 ft/s \cdot \frac{1}{5280 ft/mi}=186,000 mi/s=1.86\cdot 10^5 mi/s


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AlekseyPX

Answer:

v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

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Aleks04 [339]

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3 years ago
Read 2 more answers
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Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

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\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

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